# Problem 835 | Reactions of Continuous Beams

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$V_{1R} = \frac{2}{3} \times \frac{1}{2}(12)(150) = 600 ~ \text{lb}$ ← 1st span right support

$V_{2L} = \dfrac{400(9) + 300(4)}{12} = 400 ~ \text{lb}$ ← 2nd span left support

$V_{2R} = \dfrac{400(3) + 300(8)}{12} = 300 ~ \text{lb}$ ← 2nd span right support

$V_3 = \frac{1}{2}(12)(50) = 300 ~ \text{lb}$ ← 3rd span, left and right supports

$V_4 = 4(50) = 200 ~ \text{lb}$ ← 4th span (overhang) support

Couple Reactions

${R_1}' = 101.25 ~ \text{lb}$ first span; left support ↓, right support ↑

${R_2}' = (M_2 - M_3) / L_2 = (1215 - 661) / 12$

${R_2}' = 46.17 ~ \text{lb}$ second span; left support ↑, right support ↓

${R_3}' = (M_3 - M_4) / L_3 = [ \, 661 - 4(50)(2) \, ] / 12$

${R_3}' = 21.75 ~ \text{lb}$ second span; left support ↑, right support ↓

Thus,

$R_1 = 198.75 ~ \text{lb}$

$R_2 = 701.25 + 446.17 = 117.42 ~ \text{lb}$

$R_3 = 253.83 + 321.75 = 575.58 ~ \text{lb}$

$R_4 = 278.25 + 200 = 478.25 ~ \text{lb}$

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